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WAEC Further Mathematics 2021 Paper

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Question : 5

Total: 38

Solve (

\frac{1}{9}

^{x+2}

^{x-2}

$\frac{7}{5}$
$\frac{6}{7}$
$\frac{-7}{6}$
$\frac{-6}{7}$

Solution:

Concept:

Rewrite both sides as powers of the same base, then equate the exponents.

Explanation:

Given:

\left(\frac{1}{9}\right)^{x+2} = 243^{x-2}

Write

\frac{1}{9}

as

3^{-2}

and

243

as

3^5

.

So,

3^{-2(x+2)} = 3^{5(x-2)}

Since the bases are equal, equate the exponents:

-2(x+2) = 5(x-2)

-2x - 4 = 5x - 10

-2x - 5x = -10 + 4

-7x = -6

x = \frac{6}{7}

Answer:

x = \frac{6}{7}

Option B.

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